Consider the traditional AKM model where $$ Y_{it}=X_{it}\beta+\psi_{j(i,t)}+\epsilon_{it} $$ for $i=1,...,N$ (individual index), $t=1,...,T$ (time index), $j=1,...,J$ (firm index), and $j(i,t)$ is the firm employing worker $i$ at time $t$. For simplicity, I have ignored workers' fixed effects. $\psi_{j(i,t)}$ is called firm $j(i,t)$'s fixed effect.
By stacking observations over time, the equation above can be rewritten as $$ \underbrace{Y_i}_{T\times 1}=\underbrace{X_i}_{T\times K} \underbrace{\beta}_{K\times 1}+\underbrace{F_i}_{T\times J} \underbrace{\psi}_{J\times 1}+\underbrace{\epsilon_i}_{T\times 1} $$ where the $F(t,j)$ is $1$ if worker $i$ was employed at firm $j$ in period $t$.
Assume that:
[A1] We have a sample of observations $\{Y_i, X_i, F_i\}_{i=1}^N$ i.i.d., for $N$ large and $T$ small.
[A2] $E(\epsilon_i| X_i, F_i)=0$ for each $i=1,...,N$.
I would like to show how $\beta$ and $\psi$ are identified. I could not find any "understandable and organic" identification proof. Could you suggest one? I report here what I understood from reading/combining different sources.
My identification attempt (incomplete)
Step 1: By [A2], we have that
$$ \begin{pmatrix} E(X_i' Y_i)\\ E(F_i' Y_i)\\ \end{pmatrix}=\begin{pmatrix} E(X_i' X_i) & E(X_i' F_i)\\ E(F_i' X_i) & E(F_i' F_i)\\ \end{pmatrix} \begin{pmatrix} \beta\\ \psi \end{pmatrix} $$
Step 2: By [A1], the matrices $\begin{pmatrix} E(X_i' Y_i)\\ E(F_i' Y_i)\\ \end{pmatrix}$ and $\begin{pmatrix} E(X_i' X_i) & E(X_i' F_i)\\ E(F_i' X_i) & E(F_i' F_i)\\ \end{pmatrix} $ are consistently estimable from sample analogues. Hence, they can be treated as known in the identification exercise.
Step 3: By combining steps 2-3, it follows that if $A_{pop}\equiv \begin{pmatrix} E(X_i' X_i) & E(X_i' F_i)\\ E(F_i' X_i) & E(F_i' F_i)\\ \end{pmatrix} $ is invertible, then $\beta$ and $\psi$ are identified.
Step 4: Under which conditions is $A_{pop}$ invertible?
My understanding is that, instead of answering this question, Abowd, Kramarz, and Margolis provide necessary and sufficient conditions under which the sample analogue of the equation in Step 1 has a unique solution with respect to $(\beta,\psi)$. That is, they provide conditions under which $$ \begin{pmatrix} X'Y\\ F' Y\\ \end{pmatrix}=\begin{pmatrix} X'X & X'F\\ F' X & F'F\\ \end{pmatrix}\begin{pmatrix} \beta\\ \psi \end{pmatrix} $$ has a unique solution with respect to $(\beta,\psi)$ where $Y$, $X$, and $F$ stack $Y_i$, $X_i$ and $F_i$ over individuals. I suppose that, by law of large numbers, if the sample analogue has a unique solution, then also the population equation will have a unique solution.
Step 5: Consider separating the sample in $G$ groups satisfying the connectivity requirement discussed here.
Suppose, for simplicity, that $J=5$, $T=2$, $N=5$. Also suppose that worker 1 is employed at firm 1 in period 1 and at firm 2 in period 2; worker 2 is always employed at firm 1; worker 3 is employed at firm 3 in period 1 and at firm 2 in period 2; worker 4 is always employed at firm 3; worker 5 is employed at firm 5 in period 1 and at firm 4 in period 2. Hence, $G=2$. In group 1, there are firms 1,2,3. In group 2, there are firms 4,5. This is the example discussed in Figure 1 here.
In turn, the sample analogue of the equation in Step 1 when rearranging as discussed at p.5 of here is $$ \text{[Equation $(*)$]} \hspace{1cm}\begin{pmatrix} X'Y\\ F_1' Y\\ F_2' Y \end{pmatrix}=\underbrace{\begin{pmatrix} X'X & X'F_1 & X' F_2\\ F_1' X & F_1'F_1 & 0_{3\times 2}\\ F_2' X & 0_{2\times 3} & F_2' F_2 \end{pmatrix}}_{A_{sample}}\begin{pmatrix} \beta\\ \psi\\ \end{pmatrix} $$ where $$ F_1'F_1\equiv \begin{pmatrix} 3&0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{pmatrix}, F_2' F_2\equiv \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$$
From this point I'm lost. In particular,
(1) The authors tell us that Equation $(*)$ has a unique solution with respect to $(\beta,\psi)$ if and only if one among $\{\psi_1,\psi_2,\psi_3\}$ and one among $\{\psi_4,\psi_5\}$ are set equal to a known value (for example, $\psi_3=0$ and $\psi_4=0$). Could you help me to formally show this claim? In particular, I want to show both necessity and sufficiency.
If I look at the example above, I don't see why we need such normalisation.
I do understand that there cannot be an intercept term in $X$ in order to identify $\psi$. For simplicity, let us suppose that there is no $X$ at all.
Then, Equation $(*)$ reduces to the system $$ \begin{cases} Y_{11}+Y_{21}+Y_{22}=3\psi_1\\ Y_{12}+Y_{32}=2\psi_2\\ Y_{31}+Y_{41}+Y_{42}=3\psi_3\\ Y_{51}=\psi_4\\ Y_{52}=\psi_5\\ \end{cases} $$ that has a unique solution with respect to $\psi$ without the need of further normalising.
(2) Why, at p.10 of here, the authors go back to "expected values" rather than working with "sample means", given that we are concerned about uniqueness of solution of Equation $(*)$?
